I was given the following grammar

$ S rightarrow S (S) S mid epsilon $

First, I was asked to eliminate the left recursion. This resulted in the following:

$ S rightarrow S & # 39; $

$ S & # 39; rightarrow (S) SS & # 39; mid epsilon $

I was then asked if the grammar is LL (1). So I calculated the FIRST and FOLLOW sets as follows:

$ mathrm {FIRST} (S) = mathrm {FIRST} (S & # 39;) = {(, epsilon } $,

$ mathrm {FOLLOW} (S) = mathrm {FOLLOW} (S & # 39;) = { $, (,) } $,

I have now tried to build the LL (1) paring table $ M $ as follows:

There we have production $ S rightarrow S & # 39; $ and $ mathrm {FIRST} (S & # 39;) = {(, epsilon } $, the entry of the table $ M [ S , ( ] = S rightarrow S & # 39; $,

But since $ epsilon $ belongs in $ mathrm {FIRST} (S & # 39;) $ then for each element $ b $ in the $ mathrm {FOLLOW} (S) $,

$ M [ S , b ] = S rightarrow S & # 39; $,

That would result $ M [ S , ( ] = S rightarrow S & # 39; $,

If the entry is already filled, is this a conflict? If yes, which?

Or since the two productions are the same, can we just ignore it?

Even if the second case is the right one, is the grammar LL (1)?